Transcript Chapter 13

Chapter 13
Multiple Integrals
by Zhian Liang
13.1 Double integrals over
rectangles
Recall the definition of definite integrals of
functions of a single variable
Suppose f(x) is defined on a interval [a,b].
Taking a partition P of [a, b] into subintervals:
a  x  x   x
0
1
n 1
 x b
n
Choose the points in [ x , x ] and let  x i  x i  x i 1
i 1
i
Using the areas of the small rectangles to approximate the
areas of the curve sided echelons
and summing them, we have
n
(1)
 f ( x ) x
*
i 1
i
i
P  max{  x }
i
b
(2)
n
 f ( x ) dx  lim  f ( x ) x
*
a
P 0
i 1
i
i
Double integral of a function of two variables defined
on a closed rectangle like the following
R  [ a , b ]  [ c , d ]  {( x , y )  R | a  x  b , c  y  d }
2
Taking a partition of the rectangle
a  x  x   x
0
1
m 1
c  y  y   y
0
1
n 1
 x b
m
 y  d
n
Choosing a point
*
*
ij
ij
( x , y ) in Rij and form the
double Riemann sum
(3)
m
n
i 1
j 1
  f ( x , y )A
*
*
ij
ij
ij
(4) DEFINITION The double integral of f over the
rectangle R is defined as
m
n
i 1
j 1
 f ( x , y ) dA  lim   f ( x , y )  A
P 0
if this limit exists
*
*
ij
ij
ij
Using Riemann sum can be approximately evaluate a double
integral as in the following example.
EXAMPLE 1 Find an approximate value for the integral
 ( x  3 y ) dA , where R  {( x , y ) | 0  x  2 , 0  y  2},
2
R
by computing
the double Riemann sum with partition pines x=1 and x=3/2
*
*
ij
ij
and taking ( x , y ) to be the center of each rectangle.
Solution The partition is shown as above Figure. The area of
each subrectangle is  A  12 , ( x , y ) is the center R ,
ij
*
*
ij
ij
ij
and f(x,y)=x-3y2. So the corresponding Riemann sum is
2
2
  f ( x , y )A
i 1
j 1
*
*
ij
ij
ij
 f ( x , y )A  f ( x , y )A  f ( x , y )A  f ( x , y )A
*
*
11
11
11
*
*
12
12
12
*
*
21
21
21
 f ( 12 , 54 )  A  f ( 12 , 74 )  A  f ( 32 , 54 )  A  f ( 32 , 74 )  A
11
 
67
16
12
51 1
123 1
1
 12   139




 


16 2
16 2
16 2
  958   11 . 875
Thus we have
 ( x  3 y ) dA   11 . 875
2
R
21
22
*
*
22
22
22
Interpretation of double integrals as volumes
m
n
i 1
j 1
m
n
i 1
j 1
  v    f ( x , y )A
(5)
ij
*
*
ij
ij
ij
V    f ( x , y )A
i 1
j 1
m
n
ij
ij
*
*
ij
(6) THEOREM If f ( x , y )  0 and f is continuous on
the rectangle R, then the volume of the solid that
z  f ( x, y )
lies above R and under the surface
is
V   f ( x , y ) dA
R
R
EXAMPLE 2
Estimate the volume of the solid that lies above the square
 [ 0 , 2 ]  [ 0 , 2 ] and below the elliptic paraboloid z  16  x  2 y .
Use the partition of R into four squares and choose
( x , y ) to be the upper right corner of R .
Sketch the solid and the approximating rectangle boxes.
2
*
*
ij
ij
ij
2
Solution The partition and the graph of the function are
as the above. The area of each square is 1.Approximating
the volume by the Riemann sum, we have
V  f (1,1)  A  f (1, 2 )  A  f ( 2 ,1)  A  f ( 2 , 2 )  A
11
12
21
22
 13 (1)  7 (1)  10 (1)  4 (1)  34
This is the volume of the approximating rectangular
boxes shown as above.
The properties of the double integrals
（7） [ f ( x , y )  g ( x , y )] dA
  f ( x , y ) dA   g ( x , y ) dA
R
R
（8） cf ( x , y ) dA c  f ( x , y ) dA
R
R
（9） If
f ( x, y )  g ( x, y )
for all x , y  R ,
then
 f ( x , y ) dA   g ( x , y ) dA
R
R
R
EXERCISES 13.1
Page 837
1. 3. 15.16
13.2 Iterated Integrals
The double integral can be obtained by evaluating two
single integrals.
The steps to calculate  f ( x , y ) dA , where R  [ a , b ]  [ c , d ] :
R
Fix x to calculate
A ( x )   f ( x , y )dy
d
c
Then calculate
b
 A ( x )dx
a
with respect to y .
(1)  A ( x )dx   [  f ( x , y )dy ]dx (called iterated integral)
(2)
b
b
a
a
d
c
  f ( x , y )dy dx   [  f ( x , y )dy ]dx
b
a
d
b
c
a
d
c
(3) Similarly
  f ( x , y )dx dy   [  f ( x , y )dx ]dy
d
c
b
a
d
c
b
a
EXAMPLE 1 Evaluate the iterated integrals
3 2
2
( a )   x y dy dx
0 1
(See the blackboard)
2
3
2
(b)   x y dx dy
1
0
(4) Fubini’s Theorem If f is continuous on the
rectangle R  {( x , y ) | a  x  b , c  y  d }, then
 f ( x , y ) dA    f ( x , y )dy dx    f ( x , y )dx dy
b d
a c
d
c
b
a
R
More generally, this is true if we assume that f
is bounded on R , f is discontinuous only on
a finite number of smooth curves, and the iterated
integrals exist.
Interpret the double integral  f ( x , y ) dA as the
volume V of the solid
R
where
V   A ( x )dx
b
a
A ( x )   f ( x , y )dy
d
c
A ( x ) is the area of a cross-section of S in the
plane through x perpendicular to the x-axis.
Similarly
EXAMPLE 2
Evaluate the double integral
where
 ( x  3 y )dA
2
R
R  {( x , y ) | 0  x  2 , 1  y  2}.
(See the blackboard)
EXAMPLE 3
Evaluate
 y sin( xy ) dA
R
,where
R  [1, 2 ]  [ 0 ,  ]
Solution 1 If we first integrate with respect to x,
we get

 y sin( xy ) dA    y sin( xy )dxdy
2
0
R
1

x2
0
x 1
  [  cos( xy )]
dy

  (  cos 2 y  cos y ) dy
0
  sin 2 y  sin y 0  0
1
2

Solution 2 If we first integrate with respect to y, then

 y sin( xy ) dA    y sin( xy )dydx
2
R
1
   

2
1
1
x
0
  [
1
x
  [
1

x
2
1
2
1
 [
2
 
2
 
2
1
1
yd (cos( xy )) dx
x
2
2
sin(  x )
x
2
1
x
0
sin(  x )
x

y cos( xy ) |   cos( xy ) dy ]dx

cos(  x ) 
sin(  x )
1
0

 cos(  x )
x
dx  
2
1
dx 
1
2
x
0

sin( xy ) | ]dx
0
]dx  1
2
sin(  x )
x
2
dx  
2
1
 cos(  x )
x
dx
d sin(  x )
x
sin(  x ) 

x
2

1

2
1
sin(  x )
x
2
dx 
sin(  x ) 


x
2
1
0
EXAMPLE 4 Find the volume of the solid S that is
bounded by the elliptic paraboloid x  2 y  z  16 , the
plane y  2 and x  2 , and three coordinate planes.
2
2
Solution We first observe that S is the solid that lies
under the surface z  16  x  2 y and the above the
2
Square
R  [ 0 , 2 ]  [ 0 , 2 ].
2
Therefore,
V   (16  x  2 y ) dA
2
2
R
   (16  x  2 y ) dxdy
2
2
0
2
0

  16 x 
2
0
 
2
0

2


88
3
88
3
y
1
3
x  2y x
3
2
2

3

 4 y dy

4
3
y
2
0
 48

x2
x0
dy
If f ( x , y )  g ( x ) h ( y ) on R  [ a , b ]  [ c , d ] , then
 g ( x ) h ( y ) dA   g ( x )dx  h ( y )dy
b
a
R
d
c
EXAMPLE 5 If R  [ 0 ,  2 ]  [ 0 ,  2 ] , then
 sin x cos ydA   sin xdx  cos ydy


2
2
0
0
R
 [  cos x ] [sin y ]


2
0
 1 1  1
2
0
EXERCISES 13.2
Page 842
1(2), 6, 10, 16, 17,
13.3 Double integrals over general regions
To integrate over general regions like
R
which is bounded, being enclosed in a rectangular region R .
Then we define a new function F with domain R by
(1)
 f ( x, y )
F ( x, y )  
0
if
( x, y )  D
if
( x , y ) is in R but not in D
If F is integrable over R , then we say f is integrable
over D and we define the double integral of f over D by
(2)
where F
 f ( x , y )dA   F ( x , y )dA
D
R
is given by Equation 1.
Geometric interpretation
When
f ( x, y )  0
R
The volume under f and above D equals to that under
F and above R.
Type I regions
D  {( x , y ) | a  x  b , g ( x )  y  g ( x )}
1
2
(3) If f is continuous on a type I region D such that
D  {( x , y ) | a  x  b , g ( x )  y  g ( x )}
1
then
 f ( x , y )dA   
b
a
D
g2 ( x )
g1 ( x )
2
f ( x , y )dydx
Type II regions
(4)
(5)
D  {( x , y ) | c  y  d , h ( y )  x  h ( y )}
1

D
f ( x , y )dA 
d
 
c
h2 ( y )
2
f ( x , y )dxdy
h1 ( y )
where D is a type II region given by Equation 4
it is Type I region!

  
1
1
1 x
2x
2
2

( x  2 y ) dy dx
   xy  y
1
1


2

2
y 1  x
y2 x
2
dx
   x (1  x )  (1  x )  2 x  4 x dx
1
2
2
2
3
4
1
    3 x  x  2 x  x  1 dx
1
4
3
2
1
1


  3

2
  x
5
4
3
2

 1
x

32
15
5
x
4
x
3
x
Example 2 Find the volume of the solid that lies under the
paraboloid z  x  y and above the region D in the xy
2
2
-plane bounded by the line y  2 x and the parabola y  x .
2




Type II
Type I
D  {( x , y ) | 0  x  2 , x  y  2 x }
2
D  {( x , y ) | 0  y  4 ,
y
2
 x
y}
Type I
Solution 1
 ( x  y ) dA   (  ( x  y ) dy ) dx
2
D


2
0
y2 x
 2
3
x
y

y
dx


3  y x2

1 6
14
4
1
  ( x  x 
3
2
0
3
3

0

x
3
2
5/2
1
x ) dx  
2
3
15
y
 y x
2

2
7

y
y
x y / 2
7/2
x
8
3
2
2
4
1
6
7
4
96
y
4
x  x  x
21
0
y/2

4
0
5
5
y
6

2

216
35
0
   ( x  y ) dxdy
4
0
13
1
7
Type II4
dy  

1
3
2
x
0
2
0
D
4
2x
  ( 2 x  x  x  x ) dx
3
3
2
 ( x  y ) dA
Solution 2
2
2

1
3

y
3/2
216
35
 y
2
5/2
2

1
24
y 
3
1
2
3

y dy
Example 3 Evaluate  xydA , where D is the region bounded
by the line y  x  1 and the parabola y  2 x  6 .
D
2



D as a type I
D as a type II
Solution We prefer to express D as a type II

D  ( x, y ) |  2  y  4,
4
y 1
2
2
  
 xydA
D

1

1

2
2
y
2

4

4
2
x  y 1
4
dy
2

y ( y  1)  (


1 
1

2 
 24
 36
x 
 
y
 2 
xydxdy
2
5
4
y
y
6
5

 4y
y
4

2
3
4
y
1
2
y
2
 2y
3

 3  x  y 1
2
2
3
2
2
y
4
y
x
y
2
2
3

 3 ) dy
2
2

 8 y dy
4


2
Example 4 Find the volume of the tetrahedron bounded by
the planes x  2 y  z  2 , x  2 y , x  0 , and z  0 .
x
x

D   ( x , y ) | 0  x  1,  y  1  
2
2

D
D
Solution
1 1
   ( 2  x  2 y ) dydx
V   ( 2  x  2 y )dA
2
0 x
2
D
  2 y  xy  y
1
2
0

x
y 1 
x
2
y
x
Here is wrong in the book!
dx
2

  2  x  x 1 
1
0
x
2
  1  
   x  2 x  1 dx
1
2
0
 x x 
3
3
2
1

x
 0

1
3
x
2
2
 x
2
2
x
x

2
4
dx
Example 5 Evaluate the iterated integral
1 1
2
  sin( y )dydx
0
x
D  ( x , y ) | 0  x  1, x  y  1  ( x , y ) | 0  y  1, 0  x  y 
D as a type I
D as a type II
Solution If we try to evaluate the integral as it stands, we
are faced with the task of first evaluating  sin( y )dy . But it is
2
impossible to do so in finite terms since  sin( y )dy is not an
elementary function.(See the end of Section 7.6.) So we must
2
change the order of integration. This is accomplished by first
expressing the given iterated integral as a double integral.
  sin( y )dydx   sin( y ) dA
1 1
2
2
0 x
D
Where
D  ( x , y ) | 0  x  1, x  y  1
Using the alternative description of D,
we have
D  ( x , y ) | 0  y  1, 0  x  y 
This enables us to evaluate the integral in the reverse
order:
  sin( y )dydx   sin( y ) dA
1 1
2
2
0 x
D
   sin( y )dxdy
1
y
2
0 0
   x sin( y )  dy
1
x y
2
x0
0
  y sin( y ) dy
1
2
0


1
2
1
2
2
cos( y )

(1  cos 1)
1
0
Properties of double integrals
(6)
(7)
(8)
(9)
 [ f ( x , y )  g ( x , y )] dA   f ( x , y ) dA   g ( x , y ) dA
D
D
D
 cf ( x , y ) dA c  f ( x , y ) dA
D
D
 f ( x , y ) dA   g ( x , y ) dA
D
D
if f ( x , y )  g ( x , y ) for
all ( x , y ) in D .
 [ f ( x , y ) dA   f ( x , y ) dA   f ( x , y ) dA
D
if D  D
D1
1
 D where
2
D2
D and D
1
2
do not overlap except
perhaps on their boundaries like the following:
(10)  1dA 
(11) If m 
D
A( D )
, the area of region D.
f ( x , y )  M for all ( x , y )  D , then
mA ( D )   f ( x , y ) dA  MA ( D )
D
Example 6
U se Property 11 to estimate the integral  e
sin x cos y
D
where D is the disk with
center the region and radius 2.
 1  sin x  1, and  1  cos x  1,
Solution Since
dA ,
we have
and therefore
 1  sin x cos x  1,
e e
1
sin x cos x
e e
1
thus, using m=e-1=1/e, M=e, and A(D)=(2)2 in Property 11
we obtain
4
e
  e
D
sin x cos y
dA  4 πe
Exercises 13.3
Page 850: 7, 9, 11, 33, 35
13.4 DBOUBLE INTEGRALS IN POLAR
COORDINATE
Suppose we want to evaluate a double integral f ( x , y ) dA ,
where R is one of regions shown in the following.
R
(a)
R  {( r ,  ) | 0  r  1, 0    2 }
(b)R  {( r ,  ) | 1  r  2 ,0  
 }
Recall from Section 9.4 that the polar coordinates
of a point related to the rectangular coordinates ( r ,  )
by the equations:
r  x  y
2
2
2
x  r cos 
y  r sin 
The regions in
the above
figure are
special cases
of a polar rectangle
R  {( r ,  ) | a  r  b ,      }
Do the following partition (called polar partition)
The center of this subrectangle
R  {( r ,  ) | r  r  r , 
i 1
ij
is
ri 
*
1
1
 
( ri 1  ri )
2
j 1
i
*
j
(
j
 j)
j 1
2
  }
and the area is
A 
1
2
r  

1
2
(r  r )

1
2
( r  r )( r  r )  
ij
where
2
i
j
2
i
i
i 1
i 1
i 1
i
i 1
i
 r r 
r  r  r
r 
2
j
2
*
i
1
2
i
j
i
j
i 1
j
The typical Riemann sum is
(1)
m
n
i 1
j 1
  f ( r cos  , r sin  ) A
*
*
i
m
n
i 1
j 1
*
j
*
i
j
ij
   f ( r cos  , r sin  )r  r  
*
i
*
j
*
i
*
j
*
i
i
j
If we write g ( r ,  )  rf ( r cos  , r sin  ) , then the
above Riemann sum can be written as
m
n
i 1
j 1
  g ( r ,  ) r  
*
i
*
j
i
j
which is the Riemann sum of the double integral
  g ( r ,  )dr d 

b
a
Therefore we have
m
n
i 1
j 1
m
n
i 1
j 1
 f ( x , y ) dA  lim   f ( r cos  , r sin  ) A
R
P 0
*
*
i
j
*
i
 lim   g ( r ,  ) r  
P 0
*
i
*
j
*
j
i
j
   g ( r ,  )drd 


b
a
   f ( r cos  , r sin  )rdrd 


b
a
ij
(2) Change to polar coordinates in a double
integral If f is continuous on a polar
rectangle R given by 0  a  r  b ,      ,
where 0      2 , then

 f ( x , y ) dA    f ( r cos  , r sin  )rdrd 
R
b
a
Caution: Do not forget the factor r in (2)!
Example 1 Evaluate ( 3 x  4 y )dA , where R is the region
in the upper half-plane bounded by the circles
x  y  1, and x  y  4 .
2
R
2
Solution
2
2
2
The region R can be described as
R  {( x , y ) | y  0 , 1  x  y  4}  {( r ,  ) | 1  r  2 , 0     }
2
2
 ( 3 x  4 y )dA  0 1 ( 3 r cos   4 r sin  )rdrd 

2
2
2
2
R

   ( 3 r cos   4 r sin  )drd 
0
2
2
3
2
1
r2
  r cos   r sin  r  1 d    7 cos   15 sin  d 
15
 

  7 cos  
(1  cos 2 ) d 


2

3
4
2
0
0
0
 7 sin  

15 

15 


sin 2


2
4
2
0
15
2
Example 2 Find the volume of the solid bounded
by the xy-plane and the paraboloid z  1  x  y
2
2
D  {( x , y ) | x  y  1}
2
2
 {( r ,  ) | 0  r  1, 0    2 }
V   (1  x  y )dA
2
2
D
2
   (1  r )rdrd 
0
1
2
0
2
  d   ( r  r )dr
1
0
3
0
1
r 

r
 2


 2

4 0
2
2
4
What we have done so far can be extended to the
complicated type of region shown in the following.
(3) If f is continuous on a polar
region of the form
D  {( r ,  ) |      , h ( )  r  h ( )}
1
2
then
 f ( x , y ) dA 
D

 
h ( )
2
h1 (  )
f ( r cos  , r sin  )rdrd 
Example 3 Use a double integral to find the area enclosed
by one loop of the four-leaved rose r  cos 2
D  {( r ,  ) | 

4
 

, 0  r  cos 2 }
4
 /4
A ( D )   dA    

/4
cos 2 
0
rdrd 
D
1 
r
 2 
 /4
 



1
2
1
cos 2 
d
2
/4
 /4
0
  cos 2 d 

/4
 /4


2
/4
1  cos 4 d 
4

1
1


  sin 4

4 
4

/4


/4

8
Example 4 Find the volume of the solid that lies under the
paraboloid z  x  y , above the xy  plane, and inside
the cylinder x  y  2 x .
2
2
2
2
Solution The solid lies above the disk, whose boundary circle
D  {( r ,  ) | 

2
 

2
, 0  r  2 cos  }
 /2
2 cos 
V   ( x  y ) dA    / 2 0
2
2
r rdrd 
2
D
r 
 4 
4
 /2
 

 8
/2
 /2
0
2 cos 
d
 4   cos  d 

0
cos  d 
4
 /2
 1  cos 2 

 d
2


 8
 /2
 2  [1  2 cos 2 
0
/2
2
0
 /2
4

1  cos 4 ] d 
2
1
 /2
3

1
 2   sin 2  sin 4
8
 2

0
 3     3
 2   
2
 2  2 
Exercises 13.4
Page 856: 1, 4, 6, 7, 22